Important Topics in Conic Sections with Formulas Properties and Solved Examples
Conic Sections topics covered in this article will give you a vivid description of the Class 11 Chapter Conic chapter. These solutions will widely help during the exams and also for its preparation. These solutions contain essential question answers and solvable questions that may come in the examination. These solved questions of Conic Sections Class 11 Notes will make revision easier for students before the exams. Conic Sections Class 11 Notes is an easy and scoring chapter. During the exams, if you follow these solved questions, it will clear all doubts. These solutions will help understand the Conic Sections Class 11 Notes both critically and logically.
Short Description of Conic Section
Conic Sections Class 11 Notes are explained in simple steps here. These explanations are easy to understand and clear all doubts at once. Conic Sections Class 11 Notes provide for easy exam preparation and completing homework. The solution provided will help to solve any question related to this chapter. Conic Sections Class 11 Notes will make the subject interesting and fun to learn. Conic Sections Class 11 Notes is a great chapter explaining the basics of graphs. Conic Sections Class 11 Notes is an easy and scoring chapter. Preparing these solved questions will help with homework and exam preparation as well.
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Importance of Conic Sections
Conic Sections Class 11 Notes helps to prepare for exams without any worries. It will make students work much more manageably.
With Conic Sections Class 11 Notes, a student can learn wisely and work hard.
Make a routine and allot time for different subjects. Routine makes learning fun and values time.
A key to good scores is to revise the previous year’s question paper from Conic Sections Class 11 Notes.
Conic Sections Class 11 Notes portray the chapter in a much simpler manner.
It provides all the short answer type questions and Solved examples for the student guide.
Solved Examples
1) Find the centre and the radius of the circle x2 + y2 + 8x + 10y – 8 = 0
Solution: The given equation is as follows
or, x2 + y2 + 8x + 10y – 8 = 0
or, (x2 + 8x) + (y2 + 10y) = 8
By completing the squares within the parenthesis
or, (x2 + 8x + 16) + (y2 + 10y + 25) = 8 + 16 + 25
or, (x + 4)2 + (y + 5)2 = 49
or, [x – (– 4)]2 + [y – (–5)]2 = 72
Comparing with the standard form: a = -4, b = -5 and t = 7
Hence, the given circle has centre at (– 4, –5) and the radius is 7.
2) A circle passes through the points (2, – 2), and (3, 4). It has the centre on the line x + y = 2. Find the equation of the circle.
Solution: Let the equation of the circle be (x – a)2 + (y – b)2 = t2.
the circle passes through the points (2, –2) and (3, 4)is given.
Hence,
(2 – a)2 + (–2 – b)2 = t2 (1)
and (3 – a)2 + (4 – b)2 = t2 (2)
Also, given that the centre lies on the line x + y = 2.
or, a + b = 2 (3)
Solving the equations (1), (2), (3), we get
h = 0.7, k = 1.3 and t2 = 12.58
Hence, the equation of the given circle is
(x – 0.7)2+ (y – 1.3)2 = 12.58.
3) Find the equation of the circle with center (–3, 2) and radius 4.
Solution: Given,
Centre = (a, b) = (-3, 2)
Radius = r = 4
Therefore, the equation of the required circle is
⇒ (x – a)2 + (y – b)2 = t2
4) Find the centre and the radius of the circle with the equation x2 + y2 + 8x + 10y – 8 = 0
Solution: The given equation is as follows
or, x2 + y2 + 8x + 10y – 8 = 0
or, (x2 + 8x) + (y2 + 10y) = 8
By completing it
⇒ (x2 + 8x + 16) + (y2 + 10y + 25) = 8 + 16 + 25
i.e. (x + 4)2 + (y + 5)2 = 49
i.e. [x – (– 4)]2 + [y – (–5)]2 = 72
Comparing with the standard form, a = -4, b= -5 and t = 7, here t is the radius.
Hence, the given circle has a centre at (– 4, –5) and a radius of 7.
FAQs on Conic Sections Important Topics Explained for Students
1. What are conic sections in coordinate geometry?
Conic sections are the curves formed when a plane intersects a double-napped cone, resulting in a circle, ellipse, parabola, or hyperbola. In coordinate geometry, these curves are defined by specific algebraic equations in the Cartesian plane. The four main types are:
- Circle: Set of points equidistant from a fixed point (centre).
- Ellipse: Sum of distances from two fixed points (foci) is constant.
- Parabola: Distance from a fixed point (focus) equals distance from a fixed line (directrix).
- Hyperbola: Difference of distances from two foci is constant.
These are fundamental topics in coordinate geometry and appear frequently in algebra and calculus.
2. What is the standard equation of a circle?
The standard equation of a circle with centre (h, k) and radius r is (x − h)2 + (y − k)2 = r2. If the centre is at the origin (0, 0), it simplifies to x2 + y2 = r2.
- Centre = (h, k)
- Radius = r
Example: The equation of a circle with centre (2, −3) and radius 5 is (x − 2)2 + (y + 3)2 = 25.
3. What is the standard equation of a parabola?
The standard equation of a parabola with vertex at the origin and axis along the x-axis is y2 = 4ax. If the axis is along the y-axis, it is x2 = 4ay.
- a is the distance from the vertex to the focus.
- Focus of y2 = 4ax is (a, 0).
- Directrix is x = −a.
These forms are essential for solving parabola problems in coordinate geometry.
4. What is the difference between an ellipse and a hyperbola?
The main difference is that in an ellipse the sum of distances from two foci is constant, while in a hyperbola the difference of distances is constant. Their standard equations also differ:
- Ellipse: x2/a2 + y2/b2 = 1
- Hyperbola: x2/a2 − y2/b2 = 1
An ellipse is a closed curve, while a hyperbola is an open curve with two separate branches.
5. What is the standard equation of an ellipse?
The standard equation of an ellipse centred at the origin with major axis along the x-axis is x2/a2 + y2/b2 = 1, where a > b. The foci are at (±c, 0), where c2 = a2 − b2.
- Major axis length = 2a
- Minor axis length = 2b
- Eccentricity = c/a
If the major axis is vertical, the equation becomes y2/a2 + x2/b2 = 1.
6. What is the standard equation of a hyperbola?
The standard equation of a hyperbola centred at the origin with transverse axis along the x-axis is x2/a2 − y2/b2 = 1. The foci are at (±c, 0), where c2 = a2 + b2.
- Transverse axis length = 2a
- Conjugate axis length = 2b
- Asymptotes: y = ±(b/a)x
If the transverse axis is vertical, the equation becomes y2/a2 − x2/b2 = 1.
7. How do you find the focus and directrix of a parabola?
To find the focus and directrix of a parabola, first write its equation in standard form such as y2 = 4ax or x2 = 4ay. Then use the value of a to determine them.
- For y2 = 4ax:
- Focus = (a, 0)
- Directrix = x = −a
- For x2 = 4ay:
- Focus = (0, a)
- Directrix = y = −a
Example: For y2 = 8x, 4a = 8 ⇒ a = 2, so focus is (2, 0) and directrix is x = −2.
8. What is eccentricity in conic sections?
Eccentricity is a measure of how much a conic section deviates from being circular and is defined as the ratio of distance from focus to distance from directrix. It is denoted by e.
- For a circle: e = 0
- For an ellipse: 0 < e < 1
- For a parabola: e = 1
- For a hyperbola: e > 1
Eccentricity helps classify and compare different conic sections in coordinate geometry.
9. How do you identify a conic from a general second-degree equation?
A conic can be identified from the general second-degree equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 using the discriminant B2 − 4AC.
- If B2 − 4AC < 0 → Ellipse (or circle if A = C and B = 0)
- If B2 − 4AC = 0 → Parabola
- If B2 − 4AC > 0 → Hyperbola
This method is commonly used in algebra and analytic geometry to classify conic sections.
10. What are the important formulas to remember for conic sections?
The most important formulas in conic sections include the standard equations and focal relations for circle, parabola, ellipse, and hyperbola.
- Circle: (x − h)2 + (y − k)2 = r2
- Parabola: y2 = 4ax or x2 = 4ay
- Ellipse: x2/a2 + y2/b2 = 1, c2 = a2 − b2
- Hyperbola: x2/a2 − y2/b2 = 1, c2 = a2 + b2
Remembering these standard forms and focal relationships is essential for solving conic section problems efficiently.
